Question

The average weight of 80 randomly

selected sacks of rice is 45.54 kilos with a

standard deviation of 17 kilos. Test the

hypothesis at a 0.01 level of significance that

the true mean weight is less than 49 kilos. 2. The average length of time for students to

have their subjects controlled is 40 minutes. A

new controlling procedure using modern

computing machines is being tried. If a random

sample of 15 students has an average

controlling time of 25 minutes with a standard

deviation of 12.9 minutes under the new

system, test the hypothesis that the average

length of time to control student’s subjects is

less than 40 minutes. Use a level of

significance of 0.10 and assume the

population of controlling times to be normally

distributed.

Answer 1

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge49$

$H_a:\mu<49$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=79$ degrees of freedom, and the critical value for a left-tailed test is $t_c =-2.374482.$The rejection region for this left-tailed test is $R = \{t:t< -2.374482\}.$

The t-statistic is computed as follows:

$$t=\dfrac{45.54-49}{17/\sqrt{80}}=-1.8204$$

Since it is observed that $t = -1.8204> -2.374482=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed $df=79$ degrees of freedom, $t=-1.8204$ is $p=0.036243,$ and since $p=0.036243>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 80, at the $\alpha = 0.01$ significance level.

2. The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge40$

$H_a:\mu<40$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.10,$ $df=n-1=14$ degrees of freedom, and the critical value for a left-tailed test is $t_c =-1.34503.$The rejection region for this left-tailed test is $R = \{t:t< -1.34503\}.$

The t-statistic is computed as follows:

$$t=\dfrac{25-40}{12.9/\sqrt{15}}=-4.503469$$

Since it is observed that $t = -4.503469<-1.34503=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed $df=14$ degrees of freedom, $t=-4.503469$ is $p= 0.000248,$ and since $p= 0.000248<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 40, at the $\alpha = 0.10$ significance level.